<div class="_1l1MA">
Given the root of an n-ary tree, return the preorder traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)
Example 1:
<pre>Input: root = [1,null,3,2,4,null,5,6] Output: [1,3,5,6,2,4] </pre>
Example 2:
<pre>Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14] Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10] </pre>
Constraints:
- The number of nodes in the tree is in the range
[0, 10<sup>4</sup>]. 0 <= Node.val <= 10<sup>4</sup>- The height of the n-ary tree is less than or equal to
1000.
Follow up: Recursive solution is trivial, could you do it iteratively?
</div>
题目大意：
求n个儿子的树的前序遍历
DFS解题思路：
严格按照定义，先root，再加入儿子节点
解题步骤：
N/A
注意事项：
- root.children可能为None，所以要default成[]
Python代码：
1
2
3
4
5
6
7
8
9
10def preorder(self, root: 'Node') -> List[int]:
return self.dfs(root)
def dfs(self, root):
if not root:
return []
res = [root.val]
for child in root.children or []:
res.extend(self.dfs(child))
return res
算法分析：
时间复杂度为O(n)，空间复杂度O(1)
迭代算法II解题思路：
用stack，但不能用模板，因为太多儿子，所以类似于Iterator方法Leetcode 341 Flatten Nested List Iterator ，先把该层的儿子节点反着加入到stack，保证后加入的后遍历。
Python代码：
1
2
3
4
5
6
7
8
9
10
11
12def iterative_preorder(self, root: 'Node') -> List[int]:
if not root:
return []
res, stack = [], []
stack.append(root)
while stack:
node = stack.pop()
res.append(node.val)
for child in reversed(node.children or []):
stack.append(child)
return res
算法分析：
时间复杂度为O(n)，空间复杂度O(n)。
